A normal series without any proper refinements is called a composition series Let G = G0 normal G1 normal ... normal Gr = {1} be a normal series for G. The groups G0/G1, ... Gr-1/Gr are called composition factors of the normal series. If the series is a composition series, then the composition factors are said to be composition factors for G. A normal series for G is a chain of subgroups of G of the form G = G0 normal G1 normal ... normal Gr = {1}, Gi not equal Gi+i Assume that n not equal 4. Then An(alternating group on n elements) is simple. A group is considered cyclic if it is generated by a single element. A simple group is a group with no nontrivial normal subgroups. commutative The set residue classes mod n. A group G of order n is said to be solvable if it has a composition series whose composition factors are cyclic of prime order.
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Solvable Groups

Let us now introduce the class of groups known as solvable groups, which play a crucial role in the theory of the solution of equations in radicals. In some sense, one could say that the solvable groups are one order of magnitude more complicated than abelian groups. Before introducing the concept of a solvable group, we must define several preliminary notions. Throughout this section let G be a group of order n.

Definition 1: A normal series for G is a chain of subgroups of G of the form

G = G0 normal G1 normal G2 normal ... normal Gr = {1}, Gi not equal Gi+i,

If two normal series for G are given

G = G0 normal G1 normal ... normal Gr = {1},
G = H0 normal H1 normal ... normal Hs = {1},

then the second is said to be a refinement of the first if every Hi is a Gj for some j. The second is said to be a proper refinement of the first if it is a refinement and s > r. A normal series without any proper refinements is called a composition series.

Example 1: Let G = V4 = Z2 cross Z2. Then a normal series for G is given by

Z2 cross Z2 normal Z2 cross {1} normal {1} cross {1}.

Example 2: Let G = S3. A normal series for G is given by

S3 normal A3 normal {1}.

We will show below that the normal series in Examples 1 and 2 are composition series.

Definition 2: Let

G = G0 normal G1 normal G2 normal ... normal Gr = {1}

be a normal series for G. The groups

(1)
G0/G1, G1/G2 ... Gr-1/Gr

are called composition factors of the normal series. If the series is a composition series, then the composition factors (1) are said to be composition factors for G.

An important result about composition factors, which we will state but not prove is the theorem of Jordan and Hölder:

Theorem 3: (Jordan-Hölder): Let

G = G0 normal G1 normal ... normal Gr = {1},
G = H0 normal H1 normal ... normal Hs = {1},

be two composition series for G. Then r = s and the composition factors

Gj = Gj-1/Gj    (j = 1,...,r)

can be rearranged, so that the ith composition factor is isomorphic to

Hi = Hi-1/Hi    (i = 1,...,r).

Thus, the composition factors of G are uniquely determined up to isomorphism and rearrangement, independent of the choice of the composition series.

Recall that a simple group is a group with no nontrivial normal subgroups. Then the following result is very useful:

Proposition 4: Let

(2)
G = G0 normal G1 normal ... normal Gr = {1}

be a normal series for G This normal series is a composition series if and only if all composition factors are simple groups.

Proof: Note that (2) is a composition series if and only if for every i (1 < i < r), there does not exist a subgroup H of G such that Gi-1 normal H normal Gi, H not equal Gi-1, H not equal Gi. The condition Gi-1 normal H normal Gi is equivalent to Gi-1/Gi normal H/Gi, and the conditions H not equal Gi-1,Gi are equivalent, respectively, to H/Ginot equalGi-1/Gi,{1}. Thus, (2) is a composition series for G equivalent to for every i, the composition factor Gi-1/Gi (1 < i < r), has no nontrivial normal subgroups.

Let us consider the normal series of Examples 1 and 2. Since

Z2 cross Z2/Z2 cross {1} isomorphic to Z2,   Z2 cross {1}/{1} cross {1} isomorphic to Z2,
S3/A3 isomorphic to Z2,   A3/{1} isomorphic to Z3,

and since Z2 and Z3 are simple groups, we see that the normal series of Examples 1 and 2 are composition series. Thus the composition factors for V4 are Z2 and Z2, and the composition factors for S3 are Z2 and Z3.

Definition 5: G is said to be solvable if it has a composition series whose composition factors are cyclic of prime order.

Proposition 6: G is solvable if and only if the composition factors of G are abelian.

Proof: impication Obvious.

backwards implication Let C be a composition factor of G. Then C is a simple, abelian group. Let p be a prime dividing |C|. Then C contains an element g of order p. Since C is simple and abelian, [g] normal C, so that C = [g]. Thus, C is cyclic of prime order.

Example 3: V4 and S3 are solvable.

Example 4: Let G be abelian. Then the composition factors of G are all abelian, so that G is solvable.

Example 5: Let n > 5. Then the composition factors of the normal series

(3)
Sn normal An normal {1}

are

Sn/An isomorphic Z2,   An/{1} isomorphic to An.

Now An is simple by Abel's theorem, so that (3) is a composition series for Sn. However, for n > 5, An is not abelian. Thus, Sn is not solvable.

Example 6: Every group of prime power order is solvable

Proposition 7: Let G be a solvable group, H a subgroup of G. Then H is solvable.

Proof: Let G = G0 normal G1 normal ... normal Gr = {1} be a composition series whose composition factors are cyclic of prime order. Define

Hi = Gi intersection H    (0 < i < r).

Then H = H0 normal H1 normal ... normal Hr = {1}. Moreover, the mapping

phi: Gi-1 intersection H maps Gi-1/Gi,
phi(g) = gGi   (g element of Gi-1)

is a homomorphism with kernel Gi intersection H. Therefore, Hi-1/Hi is isomorphic to a subgroup Gi-1/Gi and therefore is either trivial or is a cyclic group of prime order. The former occurs if and only if Hi-1 = Hi. Let us omit any duplicates among H0, ..., Hr. Then the resulting chain of subgroups of H is a composition series whose composition factors are cyclic of prime order.

Proposition 8: Let G be a solvable group, H any group, f : G maps H a homomorphism. Then f(G) is a solvable group.

Proof: Let G = G0 normal G1 normal ... normal Gr = {1} be a composition series with composition factors of prime order. Then

f(G) = f(G0) normal f(G1) normal ... normal f(Gr) normal {1}.

Moreover, for i = 1,...,r,f(Gi-1)/f(Gi) is either trivial or of prime order. Thus f(G) is solvable.

Corollary 9: Let G be a solvable group, H normal G. Then G/H is solvable.

Proof: Let f : G maps G/H be the natural homomorphism. Then f(G) = G/H. The corollary now follows from Proposition 8.
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